# Cable Sizing based on I2T - 5 second boundary

In order to size a load carrying conductor after a circuit breaker, I've historically included an I2t calculation to ensure the cable used has a sufficient cross-sectional area in the event of a fault, to ensure the insulation doesn't damage.

To do this, I use the Time-Current Curve available for the breaker to calculate the worst case A2s. ( Fault Current^2 (A) x Trip Time (s) )

The Time Current curve for the part model I use shows the curve flattening at 1000s - this has historically been where I've calculated the worst case A2s.

I've recently been reading on the subject and talking to other Engineers, and I've been told that for trip times beyond 5s, the Adiabatic equation isn't really applicable.

Beyond that, if you plot a bell curve of Fault Duration against its likelihood to occur, then using a boundary of 5s would likely cover up to the 95th percentile of possible faults - which offers a good balance between safety, reliability and cost.

If the adiabatic equation is no longer applicable beyond 5s, then I obviously need to find a more appropriate way to calculate it - however I'm not convinced that a 5s boundary would cover the 95th percentile of possible faults.

Is the reasoning sound here? If so, does anyone know where this is possibly stated, to offer reasoning behind the decision?

If it is true; the 95-100th percentile of faults could cause a fire in the absolute worst case scenario, so how do you quantify the risks with that?

If not, should I be calculating up to the boundaries that the manufacturer tested for (i.e. 1000s), or is there a better argument for a lower value such as 5s?

If I've misunderstood the subject here, your help would be appreciated.

Parents
• Where the fault currents gets so low it takes that long to disconnect, then the situation is more like an overload than a fault. The simple way to deal with that is to ensure In ≤ Iz.

- Andy.

• Where the fault currents gets so low it takes that long to disconnect, then the situation is more like an overload than a fault. The simple way to deal with that is to ensure In ≤ Iz.

- Andy.

Children
•  Hi Andy,

Where the fault currents gets so low it takes that long to disconnect, then the situation is more like an overload than a fault. The simple way to deal with that is to ensure In ≤ Iz.

That sounds like the situation I'm in in relation to my posts in the thread below. An 11A earth fault current on a 10A cable (when 1mm2) with a C6 RCBO, theoretical disconnection time of a few hundred seconds. Would you apply the same logic to the few hundred second disconnection time as you would to the 1000 second above; i.e., simply ensure In <= Iz?

• An 11A earth fault current on a 10A cable (when 1mm2) with a C6 RCBO, theoretical disconnection time of a few hundred seconds.

For an earth fault there's a shock risk, so disconnection would have to occur within a much shorter time - typically 0.2s for a TT final circuit - so the long disconnection time shouldn't arise in that case. This is distinct from ensuring the protection of "undersized" live conductors for live-live faults, or where external factors (typically bonding) may reduce Zs and so has the potential to considerably increase the PFC. The 40ms disconnection times for an RCD would only protect a 1.0mm² conductor (assuming k=115) for fault currents below 575A - so only for fault currents above that would you need the OPD to provide protection. A C type is guaranteed to open by the magnetic mechanism above 10x In - so 60A for a C6 - so it sounds to me like you have all bases covered.  As you have In ≤ Iz 435.1 normally allows you to assume compliance anyway.

- Andy.